We must assume that the number of customers and the amounts that any one customers spends are all independent random variables. (This may or may not actually be a good assumption…) Then S = X1+ X2+ … + XNis a compound random variable since it satisfies the definition. We know E[N] = 10 and Var(N) = 10 since N is Poisson We know (or can derive) E[X] = 50 and Var(X) = 1002/12 since X is Uniform So using the properties for compound random variables, E[S] = E[N]E[X] = 10 * 50 = 500 Var(S) = E[N]Var(X) + (E[X])2Var(N) = 10 * 1002/12 + 502* 10 = 33,333.33 4. # 62 Let pA, pB, and pCbe the probabilities that A, B, and C win overall, respectively. We must have one winner, so pA+ pB+ pC= 1. Also, A and B are equivalent (we could just rename them) so pA= pB. We need a third equation in order to solve. Look at the probability A wins overall, conditional on the outcome of the first game. pA= P(A wins overall|wins first)*P(wins first) + P(A wins overall|loses first)*P(loses first) Let’s look at P(A wins overall|loses first) ½ the time, B will win against C, and the probability of A winning overall is 0. The other ½ the time, A will be in the exact same starting position as C – about to play the winner of the current game.

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